Copyright © . All Rights Reserved.Solution Concentration
CLASS CONCEPTS
14. Solution Concentration
Solutions are homogeneous mixtures of solute and solvent.
Solvent - the most abundant substance in a solution.
In a liquid solution, the solvent does the dissolving.
Solute - the other substance in a solution.
In a liquid solution, the solute is dissolved.
It is possible for a solution to have more than one solute, air is an example, but a solution can have only one solvent.
Solutility - The amount of solute that can be dissolved in a given amount of solvent at a given temperature.
Different substances have different .
Preparing a Saturated Salt Solution:
Sodium chloride has a solutility of 35.7 grams in 100 cm3 of COLD water. That would be 357 grams of NaCl in one liter of H2O.
Weigh out 357 grams of NaCl.
Add the salt to a 1 liter volumetric flask.
Add H2O to the graduation line and stir until dissolved.
You now have "saturated" salt water.
What is the volume of the solution?
How many moles of NaCl are in the sample?
How many particles of sodium chloride are in this sample?
How many sodium ions, Na+, are in this sample?
Concentration is a comparison of the amounts of solute and solvent.
Describing a solution as "stong" or "weak" gives some comparison of the amounts of solute and solvent, but it is only a general idea. Even the terms "dilute" and "concentrated" do not give enough information to make quantitative calculations. To be able to compare solutions quantitatively, "how much" solute and solvent must be known.
The most common units of solution concentration involves moles.
Molarity (M) = moles of solute per cubic decimeter (dm3) of solution.
Remember the following:
Volume refers to the volume of the total solution, not just the volume of the solvent.
One cubic decimeter (dm3) = 1000 cm3 = 1 liter = 1000 ml
Molarity Calculation Examples:
The dimensional analysis solution is shown for each of the following sample problems. Study how the problem is solved, understanding each step in the conversion process. When you understand, use your calculator to find each answer with the proper number of significant digits.
Note: In examples 1 and 2, the problem gives a mass of solute in a volume of solution. This may be written as "mass over volume", the form needed for molarity. The only thing needed is to convert mass to moles and volume to dm3.
1. What is the molarity of a liter of solution containing 100 g of copper (II) chloride?
2. Calculate the molarity of 100 ml of solution containing 25 g of silver nitrate.
Note: Example 3 is different. It asks "how do you prepare?" a certain volume of solution with a certain molar concentration. Even though it asks a different question, it is still a problem that converts units, therefore it is worked with dimensional analysis.
As in any dimensional analysis problem, the first step is to write down what is given in the problem. Look closely at how that information is written to begin dimensional analysis.
3. Prepare 250 ml of 0.5 M salt water.
The key difference in the two types of molarity problems above is that one type asks for a concentration and the other type provides a concentration.</span?
Making dilutions:
A solution can be made less concentrated by dilution with solvent. The number of moles of solute does not change when more solvent is added to the solution.
If a solution is diluted from V1 to V2, the molarity of that solution changes according to the equation:
M1 V1 = M2 V2
original solution 1 = diluted solution 2
The volume units must be the same for both volumes in this equation.
Dilution calculation example:
How do you prepare 100 ml of 0.40 M MgSO4 from a stock solution of 2.0 M MgSO4?
There are two solutions involved in this problem. Notice that you are given two concentrations, but only one volume.
Solution #1 is the one for which you have only concentration - the solution that is already sitting on the shelf.
Solution #2 is the one for which you have both concentration and volume - the solution that you are going to prepare.
At least until you are comfortable with this type of problem, it may be helpful to write out what numbers go with what letters in problem.
M1 = 2.0 M MgSO4 & & &
V1 = unknown
M2 = 0.40 M MgSO4 & & & V2 = 100 ml
To solve the problem, do the following:
Write the equation: M1 V1 = M2 V2
Manipulate the equation: V1 = M2 V2 / M1
Put numbers into the equation: V1 = (0.40M) (100 ml) / 2.0 M
Do the calculation: V1 = 20 ml
Describe preparing the solution as follows: Add 80 ml of distilled water to 20 ml of the 0.40 M MgSO4 solution.
Although molarity is the most common type of solution concentration used in general chemistry, there are several situations when a different comparison between solute and solvent is needed. Some of these have specialized uses, but you should be familiar with the following:
molality (m) = moles solute / Kg solvent
Colligative properties depend on the number of particles of a substance involved. Chemists do calculations dealing with changing vapor pressure, boiling point elevation, and freezing point depression in solutions. A solution concentration that compares moles of solute and kilograms of solvent is most useful in these calculations.
normality (N) = equivalents of solute / dm3 solution
Normality is a useful concentration unit to use during neutralization reactions (titrations). One mole of hydrogen ions reacts with one mole of hydroxide ions to produce water. But that doesn't mean that one mole of any acid will neutralize one mole of any base. Chemists need a unit for the amount of acid (or base) that will give one mole of hydrogen (or hydroxide) ions.
One equivalent is the amount of an acid (or base) that will give one mole of hydrogen (or hydroxide) ions.
The numerical values of normality and molarity are equal for acids and bases that give 1 equivalent of H+ or OH & per mole. For example, a solution containing 1 mole of NaOH per dm3 is 1M and also 1N. A solution containing 1 mole of H2SO4 per dm3 is 1M, but it is 2N because it contains 2 equivalents of H+ per mole.
mole fraction = the ratio of the number of moles of one substance to the total number of moles of all substances in the solution.
Mole fraction is a dimensionless quantity, like a ratio. The sum of the mole fractions of all the components in a solution must equal to ONE.
In a solution containing nA moles of solute and nB moles of solvent, the mole fraction of solute, XA, and the mole fraction of the solvent, XB, can be expressed as follows:
XA = nA / nA + nB & &
XB = nB / nA + nB
mass percent = the percent of a solution's total mass that is solute.
Many commercial solutions are labeled with mass percent. This solution concentration compares the mass of the solute to the total mass of the solution. For example, a 10% salt solution contains 10 grams of salt in each 100 grams of solution.INTRODUCTION TO MOLARITY and
solution concentrations
Brown's Chemistry - GCSE/IGCSE/GCE (basic A level)
&Online Chemical Calculations
Introducing
Molarity, volumes
and the concentration of solutions
Quantitative chemistry
calculations Help for problem solving in doing
molarity calculations from given masses, volumes and molecular/formula masses.
Practice revision questions on calculating molarity from mass, volume and
formula mass data, using experiment data, making predictions. How do we define the
concentration of a solution? How do we calculate concentration? What units do we
use for concentration? What is molarity? How do we use moles to calculate the
mass of a substance to make up a specific volume of a solution of specific
concentration? All calculation methods are fully explained with fully worked out example questions.
Online practice exam chemistry CALCULATIONS and solved problems for KS4 Science
GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level
AS/A2/IB courses. These revision notes and practice questions on how to do
molarity calculations in using solutions in chemistry and worked examples should
prove useful for the new AQA, Edexcel and OCR GCSE (9–1) chemistry science
Spotted any careless error?
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&Self-assessment Quizzes:
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Molarity, volumes
and the concentration of solutions
Appendix on
is on a separate page
See also 14.3
before proceeding in this section 11 and eventually you may need to be
familiar with the use of the apparatus illustrated above, some of which give
great accuracy when dealing with solutions and some do not.
It is very useful to be know exactly how much of a dissolved
substance is present in a solution of particular concentration or volume of a
So we need a standard way of comparing the concentrations of
solutions in.
The more you dissolve in a given volume of
solvent, or the smaller the volume you dissolve a given amount of solute in,
the more concentrated the solution.
The concentration of an aqueous solution is usually
expressed in terms of moles of dissolved substance per cubic decimetre
(reminder mole formula triangle on the right),
with units& mol
dm-3 (or mol/dm3)and this is called molarity, sometimes denoted in
shorthand as M.
1dm3 = 1 litre = 1000ml = 1000 cm3, so dividing
cm3/1000 gives dm3, which is handy to know since
most volumetric laboratory apparatus is calibrated in cm3 (or ml),
but solution concentrations are usually quoted in molarity, that is mol/dm3
(mol/litre).
Concentration is also expressed in a
'non-molar' format of mass per volume e.g. g/dm3
Equal volumes of solution of the
same molar concentration contain the same number of moles of solute i.e.
the same number of particles as given by the chemical formula.
You need to be able to calculate
the number of moles or mass of substance in an aqueous
solution of given volume and concentration
the concentration of an aqueous solution given the
amount of substance and volume of water, for this you use the equation
....& (reminder molarity formula triangle on the right),
molarity (concentration) of Z
= moles of Z / volume in dm3
This is sometimes referred to as the
mole-concentration,
and you need to be able to rearrange
this equation ... therefore ...
(1b)&moles =
molarity (concentration) x volume in dm3 and
&volume in dm3
= moles / molarity (concentration)
You may also need to know that ...
molarity x formula mass of
solute = solute concentration in g/dm3
This is sometimes referred to as the
mass-concentration,
and dividing this by 1000 gives
the concentration in g/cm3, and
concentration in g/dm3
/ formula mass = molarity in mol/dm3
both equations (2) and (3)
result from equations (1) and (4), work it out for yourself.
and to sum up, by now you should
(4) moles Z
= mass Z / formula mass of Z
(5) 1 mole = formula mass in
(6) molarity = moles/dm3
calculation Example 11.1
What mass of
sodium hydroxide (NaOH) is needed to make up 500 cm3 (0.500 dm3)
of a 0.500 mol dm-3
(0.5M) solution? [Ar's: Na = 23, O = 16, H = 1]
1 mole of NaOH = 23 + 16 + 1 = 40g
molarity = moles / volume, so
mol needed = molarity x volume in dm3
500 cm3 = 500/1000 =
mol NaOH needed = 0.500 x 0.500 =
0.250 mol NaOH
therefore mass = mol x formula
= 0.25 x 40 = 10g NaOH
calculation Example 11.2
many moles of H2SO4 are there in 250cm3 of
a 0.800 mol dm-3
(0.8M) sulphuric acid solution?
What mass of acid is in this solution?
H = 1, S = 32, O = 16]
(a) molarity = moles /
volume in dm3, rearranging equation for the sulfuric acid
= molarity H2SO4 x volume of H2SO4
= 0.800 x 250/1000 = 0.200 mol H2SO4
(b) mass = moles x formula
formula mass of H2SO4
= 2 + 32 + (4x16) = 98
0.2 mol H2SO4 x 98 =
calculation Example 11.3
potassium bromide was dissolved in 400cm3 of water.
its molarity. [Ar's: K = 39, Br = 80]
moles = mass / formula
mass, (KBr = 39 + 80 = 119)
mol KBr = 5.95/119 = 0.050
400/1000 = 0.400 dm3
molarity = moles of
solute / volume of solution
molarity of KBr
solution = 0.050/0.400 = 0.125 mol/dm3
(b) What is the concentration in grams
concentration = mass / volume, the volume
= 400 / 1000 = 0.4 dm3
concentration = 5.95 / 0.4 = 14.9 g/dm3
calculation Example
11.4 This involves calculating concentration in other ways e.g.
mass/volume units
is the concentration of sodium chloride (NaCl) in g/dm3 and g/cm3
in a 1.50 molar solution?
At. masses: Na = 23, Cl
= 35.5, formula mass NaCl = 23 + 35.5 = 58.5
since mass = mol x formula mass,
concentration = 1.5 x 58.5 = 87.8 g/dm3, and
concentration =
87.75 / 1000 = 0.0878 g/cm3
calculation
Example 11.5
A solution of calcium
sulphate (CaSO4) contained 0.500g dissolved in 2.00 dm3 of water.
Calculate the concentration in (a) g/dm3, (b) g/cm3
and (c) mol/dm3.
(a) concentration = 0.500/2.00
= 0.250 g/dm3, then since 1dm3
= 1000 cm3
(b) concentration = 0.250/1000
= 0.00025 g/cm3&& (or from 0.500/2000)
(c) At. masses:
Ca = 40, S = 32, O = 64, formula mass CaSO4 = 40 + 32 + (4 x 16) = 136
moles CaSO4
= 0.5 / 136 = 0.00368 mol in 2.00 dm3 of water
concentration CaSO4
= 0.00368 / 2 = 0.00184 mol/dm3
Molarity calculation Example 11.6
If 5.00g of sodium chloride is
dissolved in exactly 250 cm3 of water in a calibrated volumetric flask, what is the
molarity of the solution?
Ar(Na) = 23, Ar(Cl)
= 35.5, so Mr(NaCl) = 23 + 35.5 = 58.5
mole NaCl = 5.0/58.5 = 0.08547
volume = 250/1000 = 0.25 dm3
Molarity = 0. =
0.342 mol/dm3
Molarity calculation Example
There are more questions
involving molarity in
Appendix on
is on a separate page
APPENDIX on SOLUBILITY
How do you find out how soluble
a substance is in water?
Reminder: solute + solvent ==& solution
i.e. the solute is what dissolves, the
solvent is what dissolves it and the resulting homogeneous mixture is the
The solubility of a substance is the maximum
amount of it that will dissolve in a given volume of solvent e.g. water.
The resulting solution is known as a saturated
solution, because no more solute will dissolve in the solvent.
Solubility can be measured and expressed in with
different concentration units e.g. g/100cm3, g/dm3
and molarity (mol/dm3).
Solubility can also be expressed as mass of
solute per mass of water e.g. g/100g of water.
You can determine solubility by titration if the
solute reacts with a suitable reagent e.g. acid - alkali titration and it
is especially suitable for substances of quite low solubility in water e.g.
calcium hydroxide solution (alkaline limewater) can be titrated with standard
hydrochloric acid solution.
However, many substances like salts are very
soluble in water and a simple evaporation method will do which is described below
e.g. for a thermally stable salt like sodium chloride.
(1) A saturated solution is prepared by mixing
the salt with 25cm3 of water until no more dissolves at room
temperature.
(2) The solution is filtered to make sure no
undissolved salt crystals contaminate the saturated solution.
(3) Next, an evaporating dish (basin) is
accurately weighed. Then, accurately pipette 10 cm3 of the saturated
salt solution into the basin and reweigh the dish and contents.
By using a pipette, its possible to express
the solubility in two different units.
(4) The basin and solution are carefully heated
to evaporate the water.
(5) When you seem to have dry salt crystals, you
let the basin cool and reweigh it.
(6) The basin is then gently heated again and
then cooled and weighed again.
This is repeated until the weight of the dish and
salt is constant, proving that all the water is evaporated
By subtracting the original weight of the
dish from the final weight you get the mass of salt dissolved in the volume
or mass of saturated salt solution you started with.
You can repeat the experiment to obtain a
more accurate and reliable result.
(7) Calculations
By using a pipette it is possible to
calculate the solubility in two ways, expressed as two quite different
Suppose the dish weighed 95.6g.
With the 10.0 cm3 of salt
solution in weighed 107.7g
After evaporation of the water the dish
weighed 96.5g
Mass of 10.0 cm3
salt solution = 107.7 - 95.6 = 12.1g
Mass of salt in 10 cm3
of salt solution = 96.5 - 95.6 = 0.9g
Mass of water evaporated = 107.7 -
96.5 = 11.2g
(a) Expressing the solubility in grams salt
per 100 g of water
From the mass data above 0.9g of salt
dissolved in 11.2g of water
Therefore Xg of salt dissolves in 100g of
water, X = 100 x 0.9 / 11.2 = 8.0
Therefore the solubility of
the salt = 8.0g/100g water
You can scale this up to 80.0g/1000g H2O,
or calculate how much salt would dissolve in any given mass of water.
You can also express the solubility as g
salt/100g of solution.
0.9g salt is dissolved in 12.1g of
solution, Xg in 100g of solution
Therefore X = 100 x 0.9 / 12.1 = 7.4, so
solubility = 7.4g/100g solution
These calculations do not require
the original salt solution to be pipetted. You can just measure
out approximately 10cm3 of the salt solution with
10cm3 measuring cylinder, and do the experiment and
these calculations in the exactly the same way.
(b) However, if you know the exact volume of
salt solution and the mass dissolved in it, then you can calculate the
concentration in g/dm3, and if you know the formula mass of the
salt, you can calculate the molarity of the solution.
From part (a) we have 0.9g of salt in
Therefore Xg will dissolve in 1000cm3
solution, X = 1000 x 0.9 / 10 = 90g/1000 cm3
Solubility of salt = 90g/dm3
Suppose the formula mass of the salt was
200, calculate the molarity of the saturated solution.
moles salt = mass / formula mass = 90/200
= 0.45 moles
Therefore solubility of saturated
salt solution in terms of molarity = 0.45 mol/dm3
NOTE Solubility varies with temperature,
, and it usually (but not always)
increases with increase in temperature. So, in the experiment described
above, the temperature of the saturated solution should be noted, or perhaps
controlled to be saturated at 20oC or 25oC.
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